Download Solution to PHY 152 Practice Problem Set 2

Solution to PHY 152 Practice Problem Set 2
1.a. By symmetry, many of the electric forces due to individual charges on the central charge cancel. Nonzero
contributions come from the charge q ≡ +5 nC at the middle of the left boundary, and the charges q and 2q at the
lower left and upper right corner respectively. Let r = 10 cm, the net electric field at the center is
!
(q − 2q) 1
1
q
x̂ +
E=
√ (x̂ + ŷ)
4π 0 r 2
2r 2
2
√
√
1 q *
2
2 +
=
) x̂ −
ŷ
(1 −
4π 0 r 2 ,
4
4 The net electric force on the central charge is then
F = 2qE
= 2.9 × 10−5 N x̂ − 1.6 × 10−5 N ŷ
2.a. First, it is obvious that −q experiences no net electric force. By symmetry, the electric force on one of the two
+Q’s is zero if and only if the other is zero. Therefore, we can just focus on one +Q and calculate its value such
that the electric field it sees is zero. Denote the distance between −q and +Q by r,
!
−q
Q
1
+
=0
4π 0 r 2
(2r) 2
which yields Q = 4q.
2.b. For a point on the y-axis above all three charges, the electric field is
E=
1 *
q
2q
y
+
+
p
√
4π 0 , (y − 3a/2) 2 y 2 + (a/2) 2 y 2 + (a/2) 2 -
pointing in y-direction. When y a, this expression reduces to
1 *q
2q y +
+
p
4π 0 , y 2 y 2 y 2 1 3q
=
4π 0 y 2
E≈
which is the electric field of a point charge 3q.
3. Consider the situation when there is only one balloon. It stays at vertical (θ = 0) when there is no net
external force. The balloon floats because buoyance force is greater than its weight, i.e. it experiences a net force
pointing upward,
Fup = Fb − W
= ρair V g − ρHe V g
where V is the volume of the balloon and we have used Archimedes’ principle. When you try to tilt the balloon
from equilibrium by angle θ, it experiences a restoring force FR = Fup sin θ perpendicular to the direction of string.
In the case when θ is small, sin θ ≈ θ and the direction of the FR is almost horizontal. Suppose now that we have
two balloons as shown in the figure. Since they carry identical charge, the repulsive electric force FE tries to push
them apart, while FR tries to pull them inward. At equilibrium, assuming θ is small,
FE ≈ FR
Q2
1
≈ Fup sin θ
4π 0 (2L sin θ) 2
1
Q2
θ3 ≈
2
4π 0 4L ( ρair − ρHe )V g
1
3Q2
θ≈
4π 0 16L 2 ( ρair − ρHe )πr 3 g
! 1/3